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EXPLANATION OF SODIUM IONIZATIONS
By Prof. L. Kaliambos (Natural Philosopher in New Energy ) May 12 , 2015 Sodium is a chemical element with symbol Na and atomic number 11. However despite the enormous success of the Bohr model and the quantum mechanics of Schrodinger in explaining the principal features of the hydrogen spectrum and of other one-electron atomic systems, so far neither was able to provide a satisfactory explanation of ionizations of many electon atoms related to the chemical properties of atoms. Though such properties were modified by the periodic table initially proposed by the Russian chemist Mendeleev the reason of this subject of ionizations of elements remained obscure under the influence of the invalid theory of special relativity. (EXPERIMENTS REJECT RELATIVITY). It is of interest to note that the discovery of the electron spin by Uhlenbeck and Goudsmit (1925) showed that the peripheral velocity of a spinning electron is greater than the speed of light, which is responsible for understanding the electromagnetic interaction of two electrons of opposite spin. So it was my paper “Spin-spin interactions of electrons and also of nucleons create atomic molecular and nuclear structures” (2008), which supplied the clue that resolved this puzzle. Under this condition we may use this image of Sodium including the following ground state electron configuration: 1s2.2s2.2px2.2py2.2pz2.3s1. According to the “Ionization energies of the elements-WIKIPEDIA” the ionization energies (in eV) are the following: E1 = 5.13908 , E2 = 47.2864 , E3 = 71.62 , E4 = 98.91 , E5 = 138.4 , E6 = 172.18, E7 = 208.5 , E8 = 264.25, E9 = 299.864 , E10 = 1465.121, and E1 = 1648.702 . Here (- E1 ) gives the binding energy E(3s1) of the outer electron. Then, the - ( E5 + E6 + E7 ) gives the binding energy E(2px1 + 2py1 + 2pz1) of the three electrons with parallel spin , while the - ( E2 + E3 + E4 + E5 + E6 + E7 ) gives the binding energy E( 2px2 + 2py2 +2pz2) . On the other hand the -( E8 + E9) gives the binding energy E(2s2) , while the - ( E10 + E11) gives the binding energy E(1s2). See also my papers about the explanation of ionization energies of elements in my FUNDAMENTAL PHYSICS CONCEPTS and my paper of 2008. EXPLANATION OF E1 = 5.13908 eV = - E(3s1) Here the charge (-10e) of the electrons of 1s2.2s2.2px2.2py2.2pz2 screens the nuclear charge (+11e) and for a perfect screening we would have an effective Zeff = ζ = 1. However the outer electron of 3s1 penetrates the 2px2.2py2.2pz2 leading to the deformation of shells with ζ > 1. Under this condition we apply the Bohr model to write the binding energy E(3s1) = - E1 of the outer electron as E(3s1) = - E1 = - 5.13908 = (- 13.6057)ζ2/n2 Since n = 3 we get ζ = 1. 8'4' ' EXPLANATION OF - ( E5 + E6 + E7 ) = E( 2px1 + 2py1 + 2pz1) ' ' '''Here the 2px1 , 2py1 , and 2pz1 under a perfect screening of the spherical shells 1s2 and 2s2, should lead to the effective ζ = 7, because +11e -2e -2e = +7e. However the deformation of spherical shells leads to ζ > 7 . For simplicity we start with the -E7 = -208.5 eV = E(2pz1). Of course the E(2pz1) which is the binding energy of the one electron with n = 2 is based on the Bohr model. In this case we must determine the effective ζ1 > 7, because the 1s2 and 2s2 of the charge (-4e) screen the charge (+11e ) of the nucleus. Of course for a perfect screening due to the spherical shells of 1s2 and 2s2 (ζ = 7) we should write E (2pz1) = (-13.6057) ζ2/22 = (-13.6057)72/22 = - 166.67 eV However the experimental value - E7 = - 208.5 means that the 2pz1 repels the 2s2 which leads to the great deformations not only of spherical shells 1s2 and 2s2 but also of 2pz1. Thus, in fact, writing - E7 = -208.5 eV = E(2pz1) = (- 13.6057) ζ12/22 we get ζ1 = 7.83 > 7 . Then adding the second electron 2py1 one gets the binding energy E (2pz1 + 2py1) = - (E6 + E7 ). Thus applying the Bohr model we write E(2pz1 + 2py1) = 2(-13.6057)ζ22/22 = - (E6 + E7 ) = - 380.68 eV Therefore one gets ζ2 = 7.48 > 7 Then adding the third electron ( 2px1) one gets the binding energy of 3 electrons as E (2pz1 + 2py1 + 2px1) = 3(-13.6) ζ32 /22 = - (E5 + E6 + E7 ) = - 519.08 Therefore we get ζ3 = 7.13 Here ζ3 = 7.13 means that the three electrons with parallel spin (S = 1) which exert both electric and magnetic repulsions , exist at symmetrical positions providing an almost perfect screening . '''EXPLANATION OF' -(E2 + E3 + E4 + E5 + E6 + E7 ) = -736.8964 eV = E(2px2 + 2py2 + 2pz2)' ' Adding the three more electrons we get the total binding energy of 6 electrons by applying my formula of 2008 as E (2px2 + 2py2 + 2pz2) = 3+ (16.95) ζ - 4.1)/22 = - ( E2 + E3 + E4 + E5 + E6 + E7 ) = - 736.8964 eV But solving for ζ we get an effective ζ < 6 , which cannot exist. In fact, since 2px2 , 2py2 , and 2pz2 make a complete spherical shell they provide a perfect screening with ζ = 7. Thus for a correct ζ = 7 of a perfect screening we suggest that the quantum number n = 2 becomes n > 2. Under this condition for determining here the number n the above equation could be written as E (2px2 + 2py2 + 2pz2) = 3+ (16.95) 7 - 4.1)/n2 = - 736.8964 eV Then solving for n we get n = 2.23 In other words the three orbitals of paired electrons do not lead to the deformations of 1s2 and 2s2 but differ from the symmetry of (2px1+ 2py1 +2pz1) because they do not exert magnetic repulsions. Here the electric repulsions between the paired electrons of 2px2, 2py2 and 2pz2 make a complete spherical shell and lead to a perfect screening with ζ = 7. Under this condition the quantum number n = 2 becomes n = 2.23 . Note that the two electrons of opposite spin (say the 2px2) do not provide any mutual repulsion because I discovered in 2008 that at very short inter-electron separations the magnetic attraction is stronger than the electric repulsion and provides a vibration energy. However in the absence of a detailed knowledge about the mutual electromagnetic interaction between the electrons of the 2px2 or 2py2 or 2pz2 pairs today many physicists believe that it is due to the Coulomb repulsion between the two electrons of opposite spin. Under such fallacious ideas I published my paper of 2008. EXPLANATION OF -( E8 + E9 ) = E (2s2) According to the quantum mechanics the two electrons (2s2) penetrate the 1s2 shell. Thus they lead to the deformations of both 1s2 and 2s2 spherical shells giving an effective ζ > 9, because the charge (- 2e) of the two electrons of 1s2 screens the charge (+11e) of nucleus. Since n = 2 we may write ( E8 + E9 ) = - E(2s2) = - )ζ2 (+ 16.95) ζ - 4.1 / 22 Since ( E8 + E9 ) = 564.114 eV, we may write 6.8025ζ2 - 4.2375ζ - 563.089 = 0 Then, solving for ζ we get ζ = 9.41 > 9. Here 9.41 > 9. means that the repulsιοns (2s2-1s2 ) lead to the deformation of shells, because the two electrons (2s2) or (1s2) of opposite spin behave like one particle. Note that In both cases the repulsions are due to only electric forces of the Coulomb law. Whereas in the case of the three electrons of 2px1, 2py1, and 2pz1 of parallel spin (S = 1) the three electrons interact with both electric and magnetic repulsions from symmetrical positions. EXPLANATION OF - ( Ε10 + E11 ) = E( 1s2) ''' As in the case of helium the binding energy E(1s2) is due to the two remaining electrons of 1s2 with n = 1. Thus we expect to calculate the binding energy by applying my formula of 2008 for Z = 11 as E(1s2) = )112 (+ 16.95) 11 - 4.1 /12 = - 3110.06 However the experiments of ionizations give - (E10 + E11) = - 3113.823 . '''In other words one sees here that after the ionizations my formula of 2008 gives the value of 3110.06 eV which is smaller than the experimental value of 3113.823 eV. Under this condition of ionizations I suggest that n = 1 becomes n < 1 due to the fact that the ionizations reduce the electron charges and now the nuclear charge is much greater than the electron charge of the two remaining electrons. So we may write (E10 + E11) = 3113.823 eV = E(1s2) = - )112 (+ 16.95) 11 - 4.1 /n2 Then solving for n we get n = 0.9994 Category:Fundamental physics concepts